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Lecture 22 - Electrical Engineering -Part 21Slide 2© 2006 Baylor UniversityEGR 1301Electrical Engineering Topic 2:Ohm’s Law and PowerProfessor Brian Thomas SpeakingSlide 3© 2006 Baylor UniversityEGR 1301The Water AnalogyPressure in a Column of Water is equivalent to a Voltage.High Pressure Low PressureSlide 4© 2006 Baylor UniversityEGR 1301Pressure = F/AF*D = work, A*D = volume, soPressure = (F*D)/(D*A) = work/volume = energy/volumeFAreaVolumeTo hold this volume in a high pressure takes more energy (imagine a volume of water in a vacuum).DLecture 22 - Electrical Engineering -Part 22Slide 5© 2006 Baylor UniversityEGR 1301resistor a across"" voltageV =Water flow through a trough.Orifice restriction (“resistor”)High pressureLow pressureresistor a through""current I =Ohm’s LawSlide 6© 2006 Baylor UniversityEGR 1301Analogies• Battery – gives energy to _____________• Pump – gives energy to volumes of fluid• Resistor – resists the flow of electric current in a ______________ (like a wire)• Orifice – restricts the flow of fluid in a trough or pipeSlide 7© 2006 Baylor UniversityEGR 1301Circulation DuctIris (slit)Pump -mechanical energy inputHotGround10 V2 ΩHotGroundResistorIRV =Lecture 22 - Electrical Engineering -Part 23Slide 8© 2006 Baylor UniversityEGR 1301Things to Notice• Greater change in water level means greater _______________• Water drops in height by same amount that pump increases it (battery raises voltage by same amount resistor drops it)• For same pump, (constant V) increasing resistance R will _________________• Current _______ through both nodes, shallow moves faster, deep moves slowerSlide 9© 2006 Baylor UniversityEGR 1301Flashlight with Two Batteriesa) Draw schematic for this circuitb) Draw a water circuit for this circuitc) Find the current flowing in the flashlightd) Find the power delivered to the bulbSwitchΩ= 16R(press)1.5 V 1.5 VSlide 10© 2006 Baylor UniversityEGR 1301Circulation DuctIris (slit)HotGround1.5 VR=16 ΩSchematic1.5 VFan -mechanical energy input+-+-Flashlight with Two BatteriesGroundLecture 22 - Electrical Engineering -Part 24Slide 11© 2006 Baylor UniversityEGR 1301Conclude: In this case, the two batteries can be combined into one battery having a value of 3V.3 VR=16 ΩSchematicIRV =Slide 12© 2006 Baylor UniversityEGR 1301Power EquationCoulomb 1Joule 1Volt 1 =second 1Coulomb 1Amp 1 =second 1Joule 1second 1Coulomb 1*Coulomb 1Joule 1Amp) (1*Volt) (1 ==second 1Joule 11Watt =Power is Change in Energy per TimeExample: What Power is delivered by a 110 V outlet delivering 5 A?Slide 13© 2006 Baylor UniversityEGR 1301So in our flashlight example, the power delivered to the bulb is found:Lecture 22 - Electrical Engineering -Part 25Slide 14© 2006 Baylor UniversityEGR 1301One more power example:12 VR=?? ΩSchematicWP 10=IRV =RIIRIIVP2*)*(* ===IRV =RVI =RVRVVIVP2** ===Slide 15© 2006 Baylor UniversityEGR 1301Two Resistor ExampleCirculation DuctFan -mechanical energy input14 V2 Ω5 Ω2AHow much decrease in voltage occurs across each resistor?Ground14 V4 V10 VSlide 16© 2006 Baylor UniversityEGR 130114 V2 Ω5 Ω2ANote: Narrower slit (higher resistance) means larger voltage drop - the voltage drop across each resistor must obey Ohm’s Law1V+-2V+-Battery) (the VVV__21=+Conclusion:EquivalentResistance2 Ω5 ΩSo:Lecture 22 - Electrical Engineering -Part 26Slide 17© 2006 Baylor UniversityEGR 1301Summary• Water Analogy: – pressure (water height)=voltage– fluid flow = electric current• Ohm’s Law: V=IR• Power Equation: P=VI=V2/R=I2R• Batteries and resistors can be _________ using the water

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